F(x)=2x2+3x+4

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Solution for F(x)=2x2+3x+4 equation: 



(F)=2F^2+3F+4

We move all terms to the left:

(F)-(2F^2+3F+4)=0

We get rid of parentheses

-2F^2+F-3F-4=0

We add all the numbers together, and all the variables

-2F^2-2F-4=0

a = -2; b = -2; c = -4;
Δ = b2-4ac
Δ = -22-4·(-2)·(-4)
Δ = -28
Delta is less than zero, so there is no solution for the equation

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